Oozing honey through pipes
The solution below is for a very viscous fluid which has negligible inertia and large viscosity. It is wrong for water in real pipes, because it neglects the pressure drop which comes with the changing velocity of water. This term is higher order in v, but it is obviously relevant for real water pipers. I leave it, because it is an interesting exercise with a direct analogy to resistive current flow, the correct solution is at the end.
The way to do this is to note that the pressure at the divergence point is equal for all 4 pipes, and that there is a given law for pressure drop along a pipe per unit length at any a given flow rate. The answer is different depending on whether you have a fixed pressure forcing the water through the pipes (as you do in a water main system) or whether you are forcing a given volume of water through per unit time, as you suggest, and which is appropriate
when you have a large pressure drop along a very long pipe before you get to your splitter.
I will assume that the 4 pipes have a given length, and that they empty at atmospheric pressure, which I will label as 0, and that the water flow is sufficient to keep the pipes filled until near the exit point, otherwise the problem requires more information. Consider the fixed flow rate problem first. If the imposed flow rate is F units of water per second, the first equation is the mass conservation equation
$$\sum_i f_i = F $$
Where $f_i$ are the flow rates along the pipes. Phill.Zitt gave this formula, but it is not enough--- it is analogous to the current Kirchhoff Law. You also need the analog of the voltage Kirchhoff law.
The voltage law tells you that the flow rate $f_i$ is proportional to the pressure drop along pipe i. I'll call the proportionality constant the "flow conductance" $C_i$ (it's the analog of the reciprocal of resistance in an electrical circuit):
$$ f_i = C_i \Delta P $$
For the four pipes, $\Delta P$ is equal, so that
$$ f_i \propto C_i $$
and along with the sum rule, you find:
$$ f_i = {C_i f \over \sum_i C_i } $$
So the only thing you need to know are the $C_i$, just as in a resistor network.
Two pipes with flow conductances $C_1,C_2$ connected in series have a flow conductance C given by the formula:
$$ {1\over C} = {1\over C_1} + {1\over C_2}$$
For the same two pipes in parallel,
$$ C = C_1 + C_2 $$
So that conductances add in series and parallel just like the reciprocal of the resistance (the electrical conductance) in circuits. You have a problem of 4 parallel resistors connected in series to an input resistor, just like a resistor connected to 4 resistors in parallel.
For a cylindrical pipe of length L and radius R, the laminar flow profile is exactly parabolic in the radial cylindrical coordinate r:
$$ v(r) = V(1 - {r^2\over R^2}) $$
so that the total flow as a function of R is
$$ f(R) = \int_0^R v(r) 2\pi r dr = {\pi V R^2\over 2}$$
The Navier stokes equations reduce to something very simple in the laminar pipe flow case--- all the terms drop out except the viscosity term, which tells you the diffusion of momentum out of the pipe, and so the pressure drop per unit length. (see here: Is there an analytical solution for fluid flow in a square duct? )
The equation is
$$ \nu \nabla^2 v = \delta P $$
so that
$$ 2\nu {V\over R^2} = {\Delta P \over L} $$
This gives you the flow rate as a function of R and L,
$$ f = {\pi V R^2 \over 4} = {\pi R^4\over 8\nu L} \Delta P$$
so that the conductance is
$$ C(R,L) = {\pi R^4 \over 8 \nu L} $$
And this determines the flow through the i'th pipe in terms of the total flow and the geometry:
$$ f_i = {f {R_i^4\over L_i} \over \sum_k {R_k^4\over L_k}} $$
This solves the constant flow-rate problem purely geometrically.
The limit of constant flow rate is achieved when there is a long pipe feeding into the whole thing with a much larger pressure drop than the pressure drop after the split. The total flow is determined by the total conductance, which is essentially equal to the conductance of the long pipe, so no matter what you attach at the end, so long as the part at the end has much more conductance than the initial pipe.
The same problem can be solved at a fixed pressure at the divergence point, the outgoing flow is just the conductance times the shared pressure. For question 2, the issue of constant pressure or constant flow rate is essential. At constant pressure, if you attach the contraption to the side of a wide water main at high pressure, closing one pipe does nothing to the flow in the other pipes. At constant flow rate, closing pipe number 4 increases the flow through the other 3 by the factor
$$ C_1 + C_2 + C_3 + C_4 \over C_1 + C_2 + C_3 $$
For non-rigid pipes, you just need to know the R as a function of the pressure. This will be a fine approximation if the pressure drops are slow in the pipe as usual, so that the radius change slowly with length. In normal pipes, the radius doesn't change hardly at all with the pressure, so I didn't bother to calculate anything, but you can split up the pipe into slices with a radius R(P), giving a conductance, which you add according to the series rule.
Water in pipes
I will assume the flow is laminar in the pipes, but that the pipes are short, so that the pressure drop due to viscosity is negligible between the two ends. This is the correct limit for water pipes. The pressure does work on the water which is not dissipated significantly in the pipes, and comes out as kinetic energy in the water, not as heat in the pipe.
Given a pressure drop from P to atmospheric pressure 0, the water in each of the four pipes will adjust it's velocity so that the Bernoulli principle is obeyed--- the work done by the pressure is the energy gained by the water. The energy flow in a cross section of the pipe is:
$$ \int {\rho v(r)^2\over 2}v(r) 2\pi r dr $$
with the laminar profile (the flow f is as before), and this gives
$$ f {\rho V^2\over 4} $$
Where V is the velocity at the center, as before. The work done by the pressure difference at the two ends is $Pf$, so you get a version of Bernoulli's equation for laminar pipes:
$$ P + {\rho V^2\over 4} = {\rho V_0^2\over 2}$$
The velocity in the pipes are then
$$ V= \sqrt{{4P\over \rho} + {V_0^2\over 2}} $$
and they are equal. So that the flow rate in this limit (the right limit for water) is proprotional to the cross section area of the pipe, to R^2. If you have a fixed flow rate, the pressure rises to the point where the total outflow is equal to the inflow, and the water flow is partitioned according to the cross section area:
$$ f_i = {f R_i^2\over \sum_k R_k^2 }$$
This neglects the incoming velocity $V_0$, assuming the water coming out is significantly faster than the water coming in. The answer for 2 and 3 is not changed in the water case compared to the honey.
Best Answer
When you say just gravity causing the flow, do you mean that there is only gravity acting on it? No pressure difference or Friction? In that case,
Acceleration on each water molecule $= g$
Initial Velocity of each molecule, $V_I= 0$
Final velocity, $V_F= \sqrt{V_I^2 + 2gL} = \sqrt{2gL}$
For uniform acceleration, average velocity, $V_{avg}= {V_I + V_F \over 2}$
$V_{avg} = {0 + \sqrt{2gL} \over 2}$
$V_{avg} = {\sqrt2 . \sqrt{gL} \over 2}$
$V_{avg} = {\sqrt{gL} \over \sqrt2}$